Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  cout << fixed << setprecision(8);
  double n, h;
  cin >> n >> h;
  double l, r;
  double base = 1 / h;
  double mid, are, bs;
  for (double i = 1; i < n; i++) {
    l = 0.00000001;
    r = h;
    are = i / n * h / 2;
    while (r - l > 0.00000001) {
      mid = (l + r) / 2;
      bs = base * mid;
      if (mid * bs / 2 < are)
        l = mid + 0.00000001;
      else
        r = mid;
    }
    if (l * base * l / 2 == are)
      cout << l << ' ';
    else
      cout << l + 0.00000001 << ' ';
  }
  return 0;
}